# Proving hyperbolic identities [closed]

I am supposed to prove that $\tanh\ln x = \dfrac {x^2 - 1} {x^2+1}$. As far as I can tell this is not true.

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## closed as off-topic by Jonas Meyer, Adam Hughes, Claude Leibovici, saz, JohnMar 26 at 7:42

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We try to do all the details. By definition, $$\sinh(u)=\frac{e^u-e^{-u}}{2} \text{ and } \cosh(u)=\frac{e^u+e^{-u}}{2}.$$

Set $u=\ln x$, and use the fact that $e^{\ln x}=x$ and $e^{-\ln x}=\frac{1}{e^{\ln x}}=\frac{1}{x}$. We get $$\sinh(\ln x)=\frac{x-\frac{1}{x}}{2} \text{ and } \cosh(\ln x)=\frac{x+\frac{1}{x}}{2}.$$ These respectively simplify to $\frac{x^2-1}{2x}$ and $\frac{x^2+1}{2x}$. Here are the details for $\sinh(\ln x)$.

We want to simplify $\frac{x-\frac{1}{x}}{2}$. Look first at the numerator $x-\frac{1}{x}$. The second part has denominator $x$. We want the first part also to have denominator $x$, so that both parts will have a common denominator.

We have $x=\frac{x}{1}$. Multiply top and bottom by $x$. So $x=\frac{x}{1}=\frac{x^2}{x}$.

We conclude that $x-\frac{1}{x}=\frac{x^2}{x}-\frac{1}{x}=\frac{x^2-1}{x}$. Now back to $\sinh(\ln x)$. We have $$\sinh(\ln x)=\frac{x-\frac{1}{x}}{2}=\frac{\frac{x^2-1}{x}}{2}=\frac{x^2-1}{2x}.$$ A very similar calculation takes care of $\cosh(\ln x)$.

Finally, use $\tanh(u)=\frac{\sinh(u)}{\cosh(u)}$ (this is just the definition of $\tanh(u)$). We get $$\tanh(\ln x)=\frac{\sinh(\ln x)}{\cosh(\ln x)}=\frac{\frac{x^2-1}{2x}}{\frac{x^2+1}{2x}}.$$ Do the division. The $2x$ parts "cancel". Alternately, multiply top and bottom by $2x$. We get $$\tanh(\ln x)=\frac{x^2-1}{x^2+1}.$$

Comment: The derivation looks (and is) long. But that's because the steps that involve basic algebra were done in extreme detail. With some experience, one can do these steps in one's head, and the calculation can be reduced to a couple of lines.

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I don't get it, so $e^{lnx}$ is like $e^{loge}$ so I am raising e to loge? And that gives me e? –  user138246 Oct 6 '11 at 21:15
No. $x=e^{\ln(x)} =e^{\log_e(x)}$ –  Henry Oct 6 '11 at 21:23
@Jordan: No, $e^{\ln x}=x$. It is basically $\exp(\log_e x)$, but that's not the same as $\exp(\log e)$. –  anon Oct 6 '11 at 21:23
I don't understand. –  user138246 Oct 6 '11 at 21:24
You absolutely need to know the basic facts $e^{\ln x}=x$ and $\ln(e^x)=x$. The $\ln(x)$ function and the $e^x$ function are inverses of each other. You will need to use these facts over and over, both for "theory" and for practical applications. You need in fact much more. The name $\ln x$ is another name for $\log_e(x)$. So $e^{\ln x}=e^{\log_e(x)}=x$. Is that what you were asking? I could not tell for sure. –  André Nicolas Oct 6 '11 at 21:28

You have to use the fact that $\tanh t=\frac{\sinh t}{\cosh t}$ and that $\sinh t=\frac{e^t-e^{-t}}2$ and $\cosh t=\frac{e^t+e^{-t}}2$. Can you see how to proceed?

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Oh well... ${}{}{}$ –  Mariano Suárez-Alvarez Oct 6 '11 at 21:12
It is just the way I feel after a year of getting Ds and Fs no matter how hard I work. –  user138246 Oct 6 '11 at 23:57
@Jordan, obviously you're putting in the wrong kind of hard work. It's like you're spending hours and hours lifting weights and then complaining that it's not improving your piano playing. If you want to stop failing, you absolutely, unconditionally must stop treating mathematics like it was about memorizing rules that tell you exactly how to handle every problem. It just doesn't work that way. There are rules, but they don't tell you what to do. They tell you things you can do, some of which will work an some of which will not. The right thing to do is whatever happens to work. –  Henning Makholm Oct 7 '11 at 0:19
(cont.d) It looks like every time you meet any problem that you don't have an exact rule to follow for, you just lock up. Apparent you're so afraid of doing something wrong that you can't get yourself to try even simple things and see whether they work or not. But that's the only way to solve math problems: try a lot of different things and then hope one of them will turn out to work. There's no penalty for trying something that doesn't work out; throwing out a piece of paper and starting over is a free action. That's what you need to do -- it's what everyone does. –  Henning Makholm Oct 7 '11 at 0:27
(cont.d) Because if all you let yourself do is things you know will work, you will never develop any practice and intuition. You seem to think that people who are Good At Math possess some esoteric secret knowledge that lets them know instantly how to attack a problem. We don't. Perhaps we can try combinations of things a bit faster, due to experience; perhaps we're a bit more willing to try weird things that don't look likely at first. But everything we solve is by trial and error. That's the only way there is to solve new things. Rote learning simply ain't gonna give you that. –  Henning Makholm Oct 7 '11 at 0:33